Question

The hydrated salt is overheatedand the anhydrous salt thermally decomposes, one product being gas. will the reported in the hydrated salt be reported too high, too low, or be unaffected?

Answer 1

Ans. Consider the example of sodium bicarbonate pentahydrate NaHCO₃.5H₂O.

#1. Mild heating removes water of hydration form as follow-

                        NaHCO₃.5H₂O(s) -----Heat---------> NaHCO₃(s) + 5 H₂O(g)

                        Net loss per molecule = 5 H₂O

#2. Extreme heating (say, above 250.0⁰C), the compound decomposed simultaneously with dehydration as follow-

            2 NaHCO₃.5H₂O(s) -----Heat---------> Na₂CO₃(s) + CO₂(g) + 11 H₂O(g)

                        Net loss per molecule = 5.5 H₂O + CO₂

#3. So, if the compound decomposes during heating, the net loss in form of H2O would be higher (here, mass of 5.5 H₂O + CO₂) than the actual value (5 H₂O).

So, the calculated mass of water in sample would be unusually higher.

A higher mass lost in form of water would yield a higher number of moles of water lost; and thus, higher number of moles of water lost per molecules of anhydrous compound.

Therefore, the calculated water of hydration will be abnormally higher.

Say, you would get NaHCO₃.7H₂O than the actual value of NaHCO₃.5H₂O.  

In turn, you would get higher number of moles of water but lower number of moles of the dehydrated compound. So, the calculated empirical formula of the hydrated compound will contain less number of the dehydrated compound that the actual value.

Say, you would get the empirical formula of 0.5 NaHCO₃.5H₂O(s) instead of the actual empirical formula 1NaHCO₃.5H₂O(s).

Note: The empirical formula has whole number coefficient of all components. The value of 0.5 is shown here to indicate the lower value of moles of the dehydrated compound obtained.  

Result: The reported value of water of hydration will be too high.